11. The Normal distributions

Objectives (PSLS Chapter 11)

The Normal distributions

Normal distributions

The 68-95-99.7 rule

The standard Normal distribution

Using the standard Normal table (Table B)

Inverse Normal calculations

Normal quantile plots

Normal distributions

Normal curves are used to model many biological variables. They

can describe a population distribution or a probability distribution .

Normal—or Gaussian—distributions are a family of symmetrical, bell-

shaped density curves defined by a mean (mu) and a standard

deviation (sigma): N().

xx

2

2

1

2

1)(

x

exfInflection point

Inflection point

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

A family of density curves

Here means are different

( = 10, 15, and 20) while

standard deviations are the same

( = 3)

Here means are the same ( = 15)

while standard deviations are

different ( = 2, 4, and 6).

Human heights, by

gender, can be modeled

quite accurately by a

Normal distribution.

0

2

4

6

8

10

12

14

16

18

unde

r 56 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

72 o

r m

ore

Height (inches)

Per

cen

t

Guinea pigs survival times

after inoculation of a pathogen

are clearly not a good candidate

for a Normal model!

About 68% of all observations are

within 1 standard deviation

(of the mean ().

About 95% of all observations are

within 2 of the mean .

Almost all (99.7%) observations

are within 3 of the mean.

The 68–95–99.7 rule for any N(μ,σ)

Number of times σ from the center µ

All normal curves N(µ,σ) share the same properties:

To obtain any other area under a Normal curve, use either technology or Table B.

Population of young adultsN(0,1)

zStandardized bone density (no units)

What percent of young adults

have osteoporosis or osteopenia?

World Health Organization definitions of osteoporosisbased on standardized bone density levels

Normal Bone density is within 1 standard deviation (z > –1) of the young adult mean or above.

Low bone mass

Bone density is 1 to 2.5 standard deviation below the young adult mean (z between –2.5 and –1).

Osteoporosis Bone density is 2.5 standard deviation or more below the young adult mean (z ≤ –2.5).

zStandardized bone density (no units)

Women aged 70 to 79 are

NOT young adults. The mean

bone density in this age is

about −2 on the standard

scale for young adults. -6 -4 -2 0 2 4

Young adults N(0,1) Women 70-79 N(-2,1)

What is the probability that a randomly chosen woman in her 70s has

osteoporosis or osteopenia (< −1 on the standard scale)?

We can standardize data by computing a z-score:

If x has the N() distribution, then z has the N(0,1) distribution.

N(0,1)

=>

z

x

N(64.5, 2.5)

Standardized height (no units)

€

z =(x − μ)

σ

The standard Normal distribution

z (x )

A z-score measures the number of standard deviations that a data

value x is from the mean .

Standardizing: z-scores

When x is larger than the mean, z is positive.

When x is smaller than the mean, z is negative.

1 ,for

zx

When x is 1 standard deviation larger

than the mean, then z = 1.

222

,2for

zx

When x is 2 standard deviations larger

than the mean, then z = 2.

mean µ = 64.5"

standard deviation = 2.5"

height x = 67"

We calculate z, the standardized value of x:

mean from dev. stand. 1 15.2

5.2

5.2

)5.6467( ,

)(

zx

z

Given the 68-95-99.7 rule, the percent of women shorter than 67” should be,

approximately, .68 + half of (1 – .68) = .84, or 84%. The probability of

randomly selecting a woman shorter than 67” is also ~84%.

Area= ???

Area = ???

N(µ, ) = N(64.5, 2.5)

= 64.5” x = 67”

z = 0 z = 1

Women’s heights follow the N(64.5”,2.5”)

distribution. What percent of women are

shorter than 67 inches tall (that’s 5’6”)?

Using Table B

(…)

Table B gives the area under the standard Normal curve to the left of any z-value.

.0062 is the area under

N(0,1) left of z = –

2.50

.0060 is the area under N(0,1) left of

z = –2.51

0.0052 is the area under N(0,1) left of

z = –2.56

Using Excel See supplement

Area ≈ 0.84

Area ≈ 0.16

N(µ, ) = N(64.5, 2.5)

= 64.5 x = 67 z = 1

84.13% of women are shorter than 67”.

Therefore, 15.87% of women are taller than

67" (5'6").

For z = 1.00, the area

under the curve to the

left of z is 0.8413.

Tips on using Table B

Because of the curve’s symmetry,

there are two ways of finding the

area under N(0,1) curve to the

right of a z-value.

area right of z = 1 – area left of z

Area = 0.9901

Area = 0.0099

z = -2.33

area right of z = area left of –z

Using Table B to find a middle area

To calculate the area between two z-values, first get the area under

N(0,1) to the left for each z-value from Table B.

area between z1 and z2 =

area left of z1 – area left of z2

Don’t subtract the z-values!!!

Normal curves are not square!

Then subtract the

smaller area from the

larger area.

The area under N(0,1) for a single value of z is zero

The blood cholesterol levels of men aged 55 to 64 are approximately Normal with

mean 222 mg/dl and standard deviation 37 mg/dl.

What percent of middle-age men have high cholesterol (> 240 mg/dl)?

What percent have elevated cholesterol (between 200 and 240 mg/dl)?

111 148 185 222 259 296 33329 68 107 146 185 224 263 302 341

37

x zarea left

area right

240 0.49 69% 31%

200 -0.59 28% 72%

Inverse Normal calculations

You may also seek the range of values that correspond to a given

proportion/ area under the curve.

For that, use technology or use Table B backward:

first find the desired

area/ proportion in the

body of the table,

then read the

corresponding z-value

from the left column

and top row.For a left area of 1.25 % (0.0125),

the z-value is – 2.24

The lengths of pregnancies, when malnourished mothers are given vitamins and

better food, is approximately N(266, 15). How long are the 75% longest pregnancies

in this population?

?

upper 75%

The 75% longest pregnancies in this

population are about 256 days or longer.

One way to assess if a data set has an approximately Normal

distribution is to plot the data on a Normal quantile plot.

The data points are ranked and the percentile ranks are converted to z-

scores. The z-scores are then used for the horizontal axis and the actual

data values are used for the vertical axis. Use technology to obtain Normal

quantile plots.

If the data have approximately a Normal distribution, the Normal

quantile plot will have roughly a straight-line pattern.

Normal quantile plots

Roughly normal

(~ straight-line pattern)

Right skewed

(most of the data points are short survival times, while a few are

longer survival times)